JEE Main 20252 AprilMorning Shift
- Exam
- JEE Main
- Year
- 2025
- Date
- 2 April
- Shift
- Morning Shift
- Subject
- Physics
- Chapter
- Mechanical Properties of Solids
JEE Main 2025 · 2 April · Morning Shift · Physics · Mechanical Properties of Solids
A steel wire of length 2 m and Young's modulus 2.0 × 10^ 11 Nm^ -2...
A steel wire of length 2 m and Young's modulus is stretched by a force. If Poisson ratio and transverse strain for the wire are 0.2 and respectively, then the elastic potential energy density of the wire is _______ (in SI units)
Correct answer
0
Step-by-step solution
\begin{aligned} & \ell=2 \mathrm{~m} ; \mathrm{Y}=2 \times 10^{11} \frac{\mathrm{~N}}{\mathrm{~m}^2} \\ & \begin{aligned} \mu=-\frac{\left(\frac{\Delta \mathrm{r}}{\mathrm{r}}\right)}{\left(\frac{\Delta \ell}{\ell}\right)} \Rightarrow \frac{\Delta \ell}{\ell} & =\frac{1}{\mu} \times\left(\frac{\Delta \mathrm{r}}{\mathrm{r}}\right) \\ & =\frac{1}{0.2} \times\left(10^{-3}\right)\end{aligned}\end{aligned}\begin{aligned} & \Rightarrow \frac{\Delta \ell}{\ell}=5 \times 10^{-3} \\ & \begin{aligned} \mathrm{u}=\frac{1}{2} \mathrm{y} \varepsilon_{\ell}^2 & =\frac{1}{2} \times 2 \times 10^{11} \times\left[5 \times 10^{-3}\right]^2 \\ & =25\end{aligned}\end{aligned}